Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?

Question

The question arose because I wanted to understand the path of the ecliptic after reading about it here: https://johnlucey.webspace.durham.ac.uk/users/solar-year/

Is the sinusoidal path due to Earth’s axial tilt? Or is it related to mapping the path on a 2D pane? Is there a way to visualise this to aid understanding?

Answer 1

Both the equator and the ecliptic are great circles on the celestial sphere. The appearance of each on a map depends on the map projection.

In an equirectangular projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.

If the same projection is centered on the ecliptic instead, the ecliptic is a straight line and the equator is approximately sinusoidal. As uhoh's answer illustrates, neither curve is exactly a sinusoid.

Lucey provides both views too, but his ecliptic-centered map doesn't show the equator. The axes are labeled in ecliptic rather than equatorial coordinates.

He also provides a stereographic projection centered on the midday zenith at 55°N. Here both equator (red) and ecliptic (green) are mapped as circular arcs.

Answer 2

The ecliptic is a plane. It looks like a sine wave because that diagram maps the sky onto a flat plane. On the celestial sphere, the ecliptic is a great circle that crosses the celestial equator at the equinox points and which is tilted to the celestial equator by the same amount as the Earth's axial tilt (currently 23°26′12.0″).

Here's a diagram from Wikipedia that might be helpful:

Answer 3

This answer supplements the other, better, clearer answers here already.

Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?

It seems to be sinusoidal because for low inclinations the shape is roughly to sinusoidal (straight when crossing zero, has gently curved and symmetric extrema) and so we don't stop and ask what shape it is.

Equirectangular projection maps spherical coordinates $\varphi, \theta$ or lon, lat or RA, Dec (but with zero at the equator) on to cartesian $X, Y$ axes with the mind-numbingly simple transform:

\begin{align} X & = \varphi \\ Y & = \pi/2 - \theta, \\ \end{align}

but when you do that an inclined plane intersecting a unit (or celestial) sphere sphere doesn't really give you a sinusoidal wave in spherical coordinates.

Lifted from this answer to Analytical expression for the ground track of the International Space Station:

For an inclination $i$ and intersection along the $x$ axis the intersection can be described parametrically as:

\begin{align} x & = \cos t \\ y & = \sin t \ \cos i\\ z & = \sin t \ \sin i\\ \end{align}

where $t$ is the distance traveled around the circle from 0 to $2 \pi$, which you can think of as one orbit or one year, and

\begin{align} \varphi & = \arctan2(y, x)\\ \theta & = \arcsin(z).\\ \end{align}

import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
to_degs, to_rads = 180/pi, pi/180

incs = to_rads * np.arange(0, 90, 11)
t = to_rads * np.arange(-179, 180) # left out endpoints to avoid wraparound in plot

ct, st = np.cos(t), np.sin(t)

curves = []

for inc in incs:
    
    cinc, sinc = np.cos(inc), np.sin(inc)
    x, y, z = ct, st * cinc, st * sinc
    phi = np.arctan2(y, x)
    # phi = np.mod(phi + pi, twopi) - pi
    theta = np.arcsin(z)
    curves.append((inc, theta, phi))

plt.figure()
m, n = 9, 10
for i, (inc, theta, phi) in enumerate(curves):
    plt.plot(to_degs * phi, to_degs * theta)
    plt.plot(to_degs * phi[m::n], to_degs * theta[m::n], '.k')
plt.xlim(-180, 180)
plt.ylim(-90, 90)
plt.title('inclinations: 0, 11, 22, 33, 44, 55, 66, 77, 88 degrees')
plt.xlabel('RA', fontsize=12)
plt.ylabel('Dec', fontsize=12)
plt.gca().set_aspect('equal')
plt.show()